What is the integral of (ln(t)+3t^2-1) ?

The integral of (ln(t)+3t^2-1) is t^3+tln(t)-2t+C

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integral of (ln(t)+3t^2-1)

\int (ln (t) + 3 t^{2} - 1) d t

t^{3} + t ln (t) - 2 t + C

Solution steps

\int ln (t) + 3 t^{2} - 1 d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int ln (t) d t + \int 3 t^{2} d t - \int 1 d t

\int ln (t) d t = t ln (t) - t

\int 3 t^{2} d t = t^{3}

\int 1 d t = t

= t ln (t) - t + t^{3} - t

Simplify

= t^{3} + t ln (t) - 2 t

Add a constant to the solution

= t^{3} + t ln (t) - 2 t + C

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What is the integral of (ln(t)+3t^2-1) ?
The integral of (ln(t)+3t^2-1) is t^3+tln(t)-2t+C